A continuous beam ABC is supported at A, B, and C. There is a downward point load of 10 kN applied between A and B, at a distance of 4 m from A, and another downward point load of 2 kN applied between B and C, at a distance of 4 m from B. The distance between A and B is 6 m and the distance between B and C is 4 m. The problem is to determine the reaction at the support B using Castigliano's theorem.

1. Problem Description

A continuous beam ABC is supported at A, B, and C. There is a downward point load of 10 kN applied between A and B, at a distance of 4 m from A, and another downward point load of 2 kN applied between B and C, at a distance of 4 m from B. The distance between A and B is 6 m and the distance between B and C is 4 m. The problem is to determine the reaction at the support B using Castigliano's theorem.

2. Solution Steps

Let $R_A$, $R_B$, and $R_C$ be the reactions at supports A, B, and C, respectively. We need to find $R_B$.

First, consider the whole beam:

$R_A + R_B + R_C = 10 + 2 = 12$ kN

Taking moments about point A:

$6 R_B + 10 R_C = 10 \times 4 + 2 \times 10 = 40 + 20 = 60$

$6 R_B + 10 R_C = 60$

$3 R_B + 5 R_C = 30$

Castigliano's theorem states that the partial derivative of the total strain energy with respect to a force is equal to the displacement in the direction of that force. Since support B is a rigid support, its displacement is zero. We introduce a fictitious force $Q$ at point B. We set $Q = R_B$. We need to express the bending moment as a function of x and Q.

$R_A + R_B + R_C = 10 + 2$ becomes $R_A + Q + R_C = 12$

$3Q + 5 R_C = 30$

$5 R_C = 30 - 3Q$

$R_C = 6 - \frac{3}{5}Q$

$R_A = 12 - Q - (6 - \frac{3}{5}Q) = 6 - \frac{2}{5}Q$

Consider segment AB.

If $0 < x < 4$, $M = R_A x = (6 - \frac{2}{5}Q)x$

If $4 < x < 6$, $M = R_A x - 10(x-4) = (6 - \frac{2}{5}Q)x - 10(x-4) = (6 - \frac{2}{5}Q - 10)x + 40 = (-4 - \frac{2}{5}Q)x + 40$

Consider segment BC.

If $0 < x < 4$, $M = R_C x = (6 - \frac{3}{5}Q)x$

$\frac{\partial M}{\partial Q}$ for $0<x<4$ in AB is $-\frac{2}{5}x$

$\frac{\partial M}{\partial Q}$ for $4<x<6$ in AB is $-\frac{2}{5}x$

$\frac{\partial M}{\partial Q}$ for $0<x<4$ in BC is $-\frac{3}{5}x$

Using Castigliano's theorem:

$\Delta_B = \int_0^L \frac{M}{EI} \frac{\partial M}{\partial Q} dx = 0$

$\int_0^4 \frac{(6-\frac{2}{5}Q)x}{EI} (-\frac{2}{5}x) dx + \int_4^6 \frac{(-4-\frac{2}{5}Q)x+40}{EI} (-\frac{2}{5}x) dx + \int_0^4 \frac{(6-\frac{3}{5}Q)x}{EI} (-\frac{3}{5}x) dx = 0$

$EI \times 0 = \int_0^4 (6-\frac{2}{5}Q)x (-\frac{2}{5}x) dx + \int_4^6 ((-4-\frac{2}{5}Q)x+40) (-\frac{2}{5}x) dx + \int_0^4 (6-\frac{3}{5}Q)x (-\frac{3}{5}x) dx$

$0 = \int_0^4 (-\frac{12}{5} x^2 + \frac{4}{25} Q x^2) dx + \int_4^6 (\frac{8}{5}x^2 + \frac{4}{25}Qx^2 - \frac{80}{5} x) dx + \int_0^4 (-\frac{18}{5} x^2 + \frac{9}{25} Q x^2) dx$

$0 = [-\frac{4}{5}x^3 + \frac{4}{75}Qx^3]_0^4 + [\frac{8}{15}x^3 + \frac{4}{75}Qx^3 - \frac{80}{10}x^2]_4^6 + [-\frac{6}{5}x^3 + \frac{3}{25}Qx^3]_0^4$

$0 = (-\frac{4}{5}(64) + \frac{4}{75}Q(64)) + (\frac{8}{15}(216-64) + \frac{4}{75}Q(216-64) - 8(36-16)) + (-\frac{6}{5}(64) + \frac{3}{25}Q(64))$

$0 = -\frac{256}{5} + \frac{256}{75}Q + \frac{8}{15}(152) + \frac{4}{75}Q(152) - 8(20) -\frac{384}{5} + \frac{192}{25}Q$

$0 = -\frac{256}{5} + \frac{256}{75}Q + \frac{1216}{15} + \frac{608}{75}Q - 160 -\frac{384}{5} + \frac{192}{25}Q$

$0 = (-\frac{256}{5} + \frac{1216}{15} - 160 -\frac{384}{5}) + (\frac{256}{75} + \frac{608}{75} + \frac{576}{75})Q$

$0 = (-\frac{768}{15} + \frac{1216}{15} - \frac{2400}{15} -\frac{1152}{15}) + (\frac{1440}{75})Q$

$0 = (\frac{-3104}{15}) + (\frac{48}{25})Q$

$\frac{3104}{15} = \frac{48}{25}Q$

$Q = \frac{3104}{15} \times \frac{25}{48} = \frac{3104}{3} \times \frac{5}{48} = \frac{15520}{144} = \frac{970}{9} \approx 107.78$ kN

This solution looks wrong because $R_B$ is far too large.

I will try a simpler method using superposition, noting that the displacement at B must be zero.

Let $R_B$ be the reaction at B. Consider the beam AC loaded with 10 kN at 4 m from A and 2 kN at 10 m from A.

Also, consider the beam AC loaded only with $R_B$ at B.

The deflection at B due to the 10 kN and 2 kN must be equal to the upward deflection due to $R_B$.

Deflection at point B due to 10 kN:

$\delta_{B1} = \frac{P b x}{6 E I L} (L^2 - b^2 - x^2)$

Where $P = 10$, $L = 10$, $x = 6$, $a = 4$, $b = 6$

$\delta_{B1} = \frac{10 \times 6 \times 4}{6 E I (10)} (10^2 - 6^2 - 4^2) = \frac{40}{60EI}(100 - 36 - 16) = \frac{2}{3 EI}(48) = \frac{32}{EI}$

Deflection at point B due to 2 kN:

$\delta_{B2} = \frac{P a x}{6 E I L} (L^2 - a^2 - x^2)$

Where $P = 2$, $L = 10$, $x = 6$, $a = 4$, $b = 6$. (Using the other part of the formula as we are to the right of the load.)

$\delta_{B2} = \frac{2 \times 4 \times 6}{6 E I (10)} (10^2 - 4^2 - 6^2) = \frac{8}{60EI}(100 - 16 - 36) = \frac{2}{15EI}(48) = \frac{32}{5EI}$

Total downward deflection at B:

$\delta_B = \delta_{B1} + \delta_{B2} = \frac{32}{EI} + \frac{32}{5EI} = \frac{160 + 32}{5EI} = \frac{192}{5EI}$

Deflection at B due to $R_B$:

$\delta_{B3} = \frac{P a b}{6 E I L} (L^2 - a^2 - b^2)$ where a=6, b=4

$\delta_{B3} = \frac{R_B (6)(4)}{6EI (10)} (10^2 - 6^2 - 4^2) = \frac{24 R_B}{60 E I} (100 - 36 - 16) = \frac{2}{5} \frac{R_B}{EI} (48) = \frac{96}{5} \frac{R_B}{EI}$

$\frac{192}{5EI} = \frac{96}{5} \frac{R_B}{EI}$

$R_B = \frac{192}{96} = 2$ kN

3. Final Answer

2 kN